A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm?

Respuesta :

Answer:

  3.2 J

Step-by-step explanation:

It is convenient to use newtons and meters for the units when calculating work.

The spring constant (k) is ...

  k = (30 N)/(15 cm -12 cm) = (30 N)/(.03 m) = 1000 N/m

For some displacement d, the work done in stretching the spring is ...

  W = (1/2)kd^2 = (1/2)(1000 N/m)(.08 m)^2 = (1/2)(1000)(.0064) Nm

  = 3.2 J

3.2 joules of work is done stretching the spring.